1 Mean Flow Equations

Transcription

1 ME 543 Averaged Equations 23 January 2018 These notes are meant as a supplement to the text, and as an aid for the class notes, but not as an independent document. 1 Mean Flow Equations 1.1 Mean flow equations Consider the Navier-Stokes equations for constant density ρ and viscosity µ: (momentum balance ρ t U + ρ(u U = p + µ 2 U (1 (conservation of mass U = 0 (2 The viscous term can be rewritten, in index notation, as follows: with x j U j x i = µ 2 x 2 U i = µ ( Ui + U j = (2µS ij i = 1, 2, 3 j x j x j x i x j U j = 0, and S ij = 1 ( Ui + U j, the strain-rate tensor. x i x j 2 x j x i With the Reynolds decomposition, U = U + u, p = p + p, (3 plugging this into the Navier-Stokes equations gives: ρ t ( U i + u i +ρ( U j +u j ( U i +u i = ( p +p + ( 2µ( S ij +s ij x j x i x j i = 1, 2, 3 (4 with S ij = 1 2 ( U i + u i = 0, (5 x i ( Ui + U j, s ij = 1 ( ui + u j. (6 x j x i 2 x j x i Averaging Equations 4 and 5 gives, using the property of the averaging operator that it commutes with derivates: ρ t U i + ρ U j U i = p + ( 2µ S ij ρ u i u j i = 1, 2, 3 (7 x j x i x j Note that, combining Equations 5 and 8, x i U i = 0. (8 Therefore u j x j u i = u i x i = 0. (9 u i u j x j u j u i x j =0 = x j u i u j, 1

2 which was used in deriving Equation 7. Equations 7 and 8 are the same as the original Navier-Stokes equations, except for the additional term with ρ u i u j, which is call the Reynolds stress (tensor. It enters the equations in the same manner as the viscous stress tensor. It can be interpreted as the transport of mean momentum (per unit volume ρ U by the turbulent velocity field u. This is analogous to the transport of momentum by molecular motion (viscosity (see Figure 1. Figure 1: Sketch of the profiles of the downstream mean velocity of an axisymmetric jet. The profile broadens mainly due to the lateral turbulent diffusion u 1 u 2 of the mean velocity U 1. We now have the first instance of the closure problem in turbulence, i.e., we have more unknowns (10: U i, p, u i u j than equations (4. We could develop an equation for u i u j (which will be p done later, but even more unknowns will come in, e.g., u i u j u k, u i. This is a neverending process where we never have the same number of unknowns as equations, so that the x j mathematical problem is never well-posed. The closure problem is a characteristic of nonlinear stochastic processes. Consider, for example, the following nonlinear ordinary differential equation for the unknown X(t with known stochastic forcing F (t: dx dt + cx2 = F (t with c a constant. Using the Reynolds decompositions, X = X + x, F = F + f, with F known, then averaging gives, with X 2 = X 2 + x 2, d dt X + a[ X 2 + x 2 ] = F. Now we have 1 equations with 2 unknowns, X and x 2. We could form an equation for x 2, but even more unknowns will come in, and closure cannot in general be achieved. There are only two known types of problems where solutions to nonlinear stochastic equations exist. Markov processes. Turbulence is not a Markov process, but sometimes models assume that it is. If c is small, or the disturbance X is small, so that the nonlinearity is weak, and perturbation methods can be used, e.g., for weakly nonlinear waves. But turbulence is strongly nonlinear. 2

3 Note that the same closure problem arises when considering the transport by turbulence of a scalar, say Θ (e.g., temperature, concentration, etc., where Θ = Θ + θ, and where Θ satisfies: (10 Θ t + U Θ j = κ 2 Θ. (11 x j Plugging the Reynolds decomposition expressions for Θ and U i into Equation 11 and averaging gives: t Θ + U j Θ = ( κ Θ u j θ. (12 x j x j x j Note the following. The analogy with averaging the momentum equation. Again the equation looks similar to the original, but with the additional term u j θ, which is analogous to the Reynolds stress. Again u j θ can be interpreted as the transport of Θ by the turbulent motions u i, and is analogous to molecular transport (e.g., Fourier diffusion. Again a closure problem arises, where there is now one equation but four unknowns, Θ, u j θ. Equations of the type Equation 11, where the coefficient U j is a stochastic process, are sometimes referred to as parametrically driven stochastic equations, and suffer from the same closure problem, even though they are linear; however they can be recast and shown to be effectively nonlinear. 1.2 Kinetic energy of the mean flow Flow energetics are often important, e.g., In pipe flows, the power input must balance the overall kinetic energy dissipation rate (i.e., the rate of conversion of kinetic energy into internal energy. In compressible flows, the internal energy is an important dynamic quantity. In problems with gravity, e.g., for density stratified flows, potential energy is usually involved and important. In chemically reacting flows, chemical and internal energy are often exchanged with kinetic energy. In turbulence, the kinetic energy of the mean flow, 1 2 ρ U i 2, and of the turbulence, 1 2 ρ u2 i, often must be considered (see Figure 2. The equation for the kinetic energy of the mean flow, 1 2 ρ U i 2, can be obtained by multiplying the mean momentum equation (Equation 7 by U i and contracting the indices (that is, taking the x 2 j 3

4 dot product of the mean momentum equation with U. The result is, after some manipulations: ( t + U j ( ρ U i 2 x j 2 time-rate-of-change of mean flow KE following the mean flow = { } U j p + 2µ U i S ij ρ U i u i u j x j spatial redistribution of mean flow KE due to pressure forces, viscous stresses, Reynolds stresses 2µ S i j S ij loss of mean flow KE into internal energy + ρ u i u j S ij loss (gain of mean flow KE to (from turbulence KE. (13 Figure 2: Plot of the direction of transfer of kinetic energy in turbulence, from the mean flow kinetic energy, into both turbulent kinetic energy (the principal loss and internal energy, and from turbulent kinetic energy into internal energy. In developing this equation, the following result was used. Since S ij and u i u j are symmetric tensors, then S ij U i x j = S ij ( S ij + Ω ij = S ij S ij since S ij Ω ij = 0 because Ω ij is an anti-symmetric tensor, where Similarly, u i u j U i x j = u i u j S ij. Note the following. Ω ij = 1 2 ( Ui U j. x j x i Integrating the spatial redistribution term over the entire problem volume, using the divergence theorem, with n as the unit outward normal: V x j { } { } U j p +2µ U i S if ρ U i u i u j dv = U j p +2µ U i S ij ρ U i u i u j n j da A The integral on the right-hand-side is 0 for at least three different conditions. 1. no-slip boundary conditions, where U, u = 0 at the boundary; 2. periodic boundary conditions; 3. U, u 0 as x. 4

5 Therefore this term usually acts to redistribute energy in space, without generating energy or dissipating energy. There are some cases of boundary motion, however, (surface wave motion, motion of a piston where boundary work may be important. The term 2µ S ij S ij is always negative. There is the same term, but opposite sign, in the equation for the average internal energy; this term represents the dissipation rate of mean flow kinetic energy. The term u i u j S ij is almost always negative, so that energy is lost from the mean flow and gained by the turbulence, since the same term of opposite sign appears in the turbulence kinetic energy equation (which will come up shortly. 1.3 Estimates of the importance of various terms in the mean flow KE equation It is useful to estimate the importance of the various terms in the mean flow kinetic energy equation. This will allow us to see what physics are of importance, and to simplify the equation, using some experimental results and theory. Generally rough estimates of the various terms in the mean kinetic energy equation can be made in the following manner (for simple free shear flows such as jets, wakes, and mixing layers, see Figure 3; an axisymmetric turbulent jet is used in the example. Figure 3: Sketch of a mean velocity profile in a jet, U 1 (x 2, and a turbulent r.m.s. velocity fluctuation profile, u 2 1 1/2 (x 2, showing the estimates for the velocity scale U and length scale l u. We will call the velocity fluctuation about the mean, i.e., u i = U i U i, the turbulence velocity. Note that it cannot be characterized by its average, which is 0; so it is usually characterized by its root-mean-square (r.m.s., i.e., u 2 i 1/2. The characteristic velocity U for the turbulent velocities is taken to be defined by: 1 3 u2 i U 2. max So our rough estimate for u 2 i to be U 2, and we write this as u 2 i U 2. In addition it is known experimentally that U i 3 u 2 max i 1/2 max, so we take as our estimate for U i to be U i U as well. Furthermore, we define l u to be an estimate for the differential scale of any averaged quantity; for example for the axi-symmetric jet, l u could be taken to be the jet diameter. More precisely, l u could be taken to be the spatial integral scale of u. We will then estimate derivatives as, for example, U i U. Regarding the off-diagonal terms in the Reynolds stress, e.g., u 1 u 2, note x j l u 5

6 that, for any i, j, e.g., (1,2 ρ 12 = u 1 u 2 u 2 1 1/2 u 2 2 1/2 1 so we will estimate any of the components of u i u j as U 2. To estimate the mean pressure p we will use the dynamic pressure, which is a useful estimate for incompressible flows using arguments, e.g., from Bernoulli s equation. Therefore p ρu 2. Finally differential time will be estimated as an advective time, i.e., t l u /U. When these estimates are used, we obtain the following estimates of each term in the averaged momentum equation, Equation 7. ( t + U j x j 1 2 ρ U i 2 ρu 3 /l u + ρ u i u j S ij + ρu 3 /l u = x j { U j p ρ U i u i u j } ρu 3 /l u {2µ U i S ij } 2µ S ij S ij, (14 x j µu 2 /l 2 u =ρu ρu 3 /l u 1/Re lu 3 /l u µ/ρul u=ρu 3 /l u 1/Re lu where Re lu = ρul u /µ, and we expect that Re lu 1. All the terms are roughly of the same order of magnitude (ρ U 3 /l u except for the viscous terms, which are of the order ρ U 3 /l u (1/Re lu. For most turbulent flows, except near boundaries, where both U and l u come very small, Re lu is very large (of the order of thousands, millions so that the viscous terms in this equation are unimportant. (We will find, however, that certain viscous terms, e.g., the dissipation rate term in the turbulent kinetic energy equation, are not small. In particular, viscous diffusion turbulent diffusion {2µ U i S ij } x j {ρ U i u i u j } x j 1 Re lu 1, and viscous dissipation rate loss of (1/2 Ui 2 to (1/2/ u2 i 2µ S ij S ij ρ u i u j S ij 1 1. Re lu Therefore, except near boundaries, turbulent diffusion vastly dominates viscous diffusion, and the conversion of mean flow kinetic energy into turbulent kinetic energy vastly dominates the conversion of mean flow kinetic energy into internal energy. This implies that the properties of the mean flow, ρ U and p, are independent of the Reynolds number since, neglecting the viscous terms, viscosity no longer shows up in the mean momentum equation. This is an example of Reynolds number similarity, a topic which will be taken up after we have discussed the turbulence energy equation. These results on the dominance of turbulent diffusion and dissipation rate are part of the reason for the technological importance of turbulence. 1.4 More on the Closure Problem This is a thumbnail sketch; a much more indepth discussion is in Chapters 10 and 11 of the text. Consider again the averaged Navier-Stokes equations assuming that ρ and µ are constant. ρ t U i + ρ U j U i = p + ( 2µ S ij ρ u i u j x j x i x j i = 1, 2, 3 (15 x i U i = 0. (16 6

7 As mentioned, Equation 15 is almost the same as the original equation for U, except for the addition of the Reynolds stress term ρ u i u j, so that the equation is not closed. Some strategies to deal with the closure problem are as follows. Mean momentum equation closure (mean field closure. The idea here is to assume a relationship between u i u j and U i, which adds 6 equations, but is completely ad hoc, i.e., not justifiable mathematically. One this is done, the equations are closed. The model for the u i u j is generally based on the analogy with the kinetic theory of gases, and as we shall see is wrong in principle. It sometimes works, however, in practice, mainly for very simple flows. Reynolds stress closure. Here the equations for u i u j are derived. Now we are left with making closure assumptions for terms of the form: u i u j u k, p u i, p s ij, 2µ s ik s jk. Generally models are put forth for the first three of these terms, and the last term is related to the mean turbulent dissipation rate ɛ = 2ν s ij s ij. An equation is also developed (derived for ɛ. More closure assumptions are also required to treat the equation for ɛ. The result is a very large number of unjustifiable assumptions, which some people suggest is getting into very sophisticated curve fitting. Intermediate course, k-ɛ (or k-ω closure, where k = (1/2 u 2 i. Here the equations for U i, k, and ɛ are solved. Assumptions are made regarding: 1. how u i u j is related to U i, k, and ɛ, 2. and how u i u j u k is related to U i, k, and ɛ; 3. and other assumptions. Consider mean momentum closure, although the following ideas will also apply to other closures. The simplest idea is to follow Boussinesq and to treat the Reynolds stress, ρ u i u j in analogy to a viscous effects, i.e., ρ u i u j = 2µ T S ij, or u i u j = 2ν T S ij with ν T = µ T /ρ, where ν T is a turbulent or eddy viscosity. The value of ν T would be expected to be very large compared to viscous diffusion, since u i u j 2ν S ij U 2 ν(u/l u = Ul u ν = Re lu = ν T ν 1. Furthermore, as we will discuss below, ν T would not be expected to be constant, since the turbulence would be expected to be stronger in one region than another, i.e., ν T should depend on the flow field, and not on the fluid properties, as does ν. The expression for the Reynolds stress above must be modified for consistency since, if we contract the indices (form the trace, ρ u i u j ρ u 2 i = ρ( u u u 2 3, whereas S ij S ii = 1 ( Ui + U i = 0 from continuity. 2 x i x i So the Boussinesque assumption is modified to: u i u j = 1 3 u2 k δ ij 2ν T S ij 7

8 which reduces properly when the subscripts are contracted. In terms of the deviatoric (having 0 trace, anisotropic (is 0 if u i u j has only diagonal elements part of u i u j, this is: a ij = u i u j 2 3 kδ ij = 2ν T S ij. (17 When Equation 17 is plugged into Equation 15, there results: ρ t U i + ρ U j U i = ( p + 1 x j x i 3 u2 k + ( (2µ + ρν T S ij x j i = 1, 2, 3 (18 Therefore, (1/3 u 2 k acts as a turbulent pressure, and the role of the Reynolds stress model as a viscous stress is made more clear. In doing this we have gone from the new dependent variables (unknowns u i u j to the unknown ν T. The question is, how do we model ν T? There is a lengthy discussion of this in Chapter 10 of the text; it will only briefly be discussed here. The assumptions behind much of the transport modeling are very similar for all turbulent transport modeling. Given the variable Φ, which could be momentum (U, turbulent kinetic energy (1/2ρ u 2 k, temperature (Θ, etc., and making the Reynolds decomposition for Φ, i.e., Φ = Φ + φ, then the assumption is where K φ is often related to ν T as u i φ = K φ Φ x i (19 K φ = ν T P r T, where P r T is called a turbulent Prandtl number for Φ, and is often assumed to be O(1, since turbulence diffuses all the properties at about the same rate. So the question comes down to how to model ν T Mixing length models and more Much of turbulence modeling takes a mixing length approach, and it is useful to see its tie to the physics, and its limitations. Note that, first of all, in using the model given by Equation 17, i.e., that a ij = 2ν T S ij, this implies that a ij and S ij have the same tensorial properties, e.g., the same principal axes. It is found from one of the computer problems that the principal axes of a ij and S ij are not nearly aligned, which is a clear weakness in the model. The modeling for ν T is usually based upon arguments making the analogy between turbulence and the kinetic theory of gases. The latter, for example, rigorously predicts the viscosity ν of a simple gas, given certain properties of the gas. Specifically, the prediction is: ν = µ/ρ = αλ g c, where (20 α is a constant of order 1, and its value is determined by the theory; λ g is a length scale related to the mean free path of the gas (the average distance between molecules; and c is related to the sound speed of the gas. 8

9 A principal assumption required in the derivation of Equation 20 is that λ g l u 1, (21 where l u is a characteristic length scale of the fluid motion, e.g., a boundary layer thickness, or a jet width. Note that Equation 21 is closely related to the continuum approximation for a fluid. In analogy with the kinetic theory of gases, it is assumed for turbulence that, for example, in a turbulent shear layer (see Figure 4, Figure 4: Sketch of a turbulent shear layer with U defined in terms of the mean velocity difference, and λ T in terms of the shear layer thickness. u 1 u 2 = ν T U 1 x 2 where the term U 2 is assumed to be small in this example. x 1 In analogy with the kinetic theory of gases, the turbulent viscosity is assumed to be: ν T = α T λ T u T. Note that now u T and λ T are determined from the flow field (not from the fluid. α T is usually determined by curve-fitting (this is modeling, not a theory. In the mean field closure, u T is related to U 1 (see Figure 4. λ T is related to some specific feature of U 1, like the shear layer thickness or the jet width, and is sometimes called the mixing length. In the k-ɛ modeling, u T and λ T are related to k and ɛ in the following way: u T k 1 2, λt k 3 2 /ɛ, so νt k 2 /ɛ. Note the following regarding the modeling for ν T. As mentioned above, the tensorial properties of a ij and S ij are found experimentally to be very different. The requirement that λ T /l u 1 is not met. Instead of molecular collisions, the transport mechanism is the eddies in the turbulent flow. And in fact, experimentally it is found that λ T is of the order of the integral scale, and so λ T l U, and so λ T /l u 1, and the theory does not go through. Therefore the model is wrong in principal. 9

10 The turbulent viscosity is determined by the fluid motion, and not by the fluid itself, the latter which is the case for the laminar viscosity. The same arguments can be made for other transported quantities, e.g., heat, chemical species, kinetic energy, etc. Even with these flaws, the model works surprisingly well for simpler flows As stated by Richard Feyman (1996, even wrong theories may help in designing machines. And John Lumley (1992, one of the premier researchers on turbulence in the second half of the last century, said in our present state of understanding, these simple models will be based, in part on good physics, in part on bad physics, and in part on shameless phenomenology. Later in his career Lumley worked extensively on modeling, and at first jokingly considered publishing under an assumed name. 1.5 Turbulent kinetic energy equation Development and analysis of the turbulent kinetic energy equation will give us some significant insights into the behavior of, and the relationship between important length scales in, turbulent flows. We start with the equation for the turbulence fluctuating velocity u i. This can be obtained by subtracting the equation for the mean velocity U i (Equation 7 from the original momentum equation for U i = U i + u i (Equation 1 giving: ρ t u i + ρ U j u i + ρu j U i + ρu j u i = p + 2µ s ij + u i u j. (22 x j x j x j x i x j x j The turbulent kinetic energy equation can be obtained by multiplying Equation 22 by u i, summing over i, and averaging, giving, ( t + U j 1 x j 2 ρ u2 k = p u i ρ u i u j U i u 2 i ρ u j + 2µ u i s ij, x i x j x j 2 x j where u i = 0 has been used, and, with u i s ij = u i u i s ij s ij = u i s ij s ij s ij, x i x j x j x j x j ( t + U j x j 1 2 ρ u2 k time rate of change of turbulent KE following the mean flow 2µ s ij s ij average loss of turbulent KE to internal energy Note the following. = { p u 2 k u i ρ u i + 2µ u j s ij x i 2 spatial redistribution due to pressure forces, transport by turbulence, and transport by viscosity ρ u i u j S ij gain (sometimes loss of turbulent KE from (sometimes to the mean flow KE. (23 The term ρ ɛ = 2µ s ij s ij is called the turbulent viscous dissipation rate; it is always positive and its effect is always a loss of turbulent kinetic energy into internal energy. A term of opposite sign appears in the averaged internal energy equation. As with the equation for the mean flow kinetic energy, the spatial redistribution term is normally conservative, as energy is moved spatially, but without any gain or loss of total energy. 10

11 The term ρ u i u j S ij appears in the equation for the mean flow kinetic energy (Equation 13, but with opposite sign, and usually represents a transfer of kinetic energy from the mean flow to the turbulence. For this reason it is sometimes called the production term. As we will see in the scaling analysis of this equation, the turbulent dissipation rate ɛ is very significant in the dynamics of turbulent flows, and in the modeling, e.g., in the k-ɛ modeling. 1.6 Scaling analysis of the turbulent kinetic energy equation Some very useful information can be learned from scaling analysis of the turbulent energy equation, but not without some controversy. The following choices will be made to estimate the magnitude of the terms in the equation. u 2 i U 2, U i U, x j ( 1 l u, t ( U l u, the same estimates used in the mean flow kinetic energy equation. In addition, Note that u j s ij = u j 1 2 u i u j u k U 3, p ρ U 2 so p u i ρ U 3. ( ui + u j x j x i = 1 u i u j + 1 u 2 j, so, using the scaling above, 2 x j 2 x i 2 u j s ij U 2 l u. The Taylor spatial microscale λ is defined here as ( 2 u1 x 1 = u2 1 λ 2, (24 so that s ij s ij U 2 λ 2. Using these estimates, the terms in the turbulent energy equation scale as follows. ( t + U j x j 1 2 ρ u2 k ρ U 3 /l u 2µ s ij s ij µ(u 2 /λ 2 =(ρ U 3 /l u(l 2 u /λ2 (µ/ρ Ul u p u 2 } k u i ρ u i 2 ρ U 3 /l u { } 2µ u j s ij x j = { x i + µ(u 2 /l 2 u =(ρ U 3 /l u(µ/ρ Ul u ρ u i u j S ij ρ U 3 /l u. (25 Again, Re l = ρ Ul u /µ will assumed to be large. All the terms are of order ρ U 3 /l u, and hence generally in balance, expect for the following terms. 1. Compared to the others, the viscous diffusion term behaves as 1/Re l, which implies that it is very small and can be neglected, except possibly at boundaries, or flow low Reynolds number flows. ρ U 2 ( 2 lu 1 2. The dissipation rate term behaves as:. Kinetic energy dissipation is a principal effect of turbulence, which implies that this term should be of the same order as the l u λ Re l others. This implies the following. 11

12 ( λ lu = O(Re 1/2 l as Re l, giving the Reynolds number dependence of the ratio of the microscale to the large (integral scale. (bandwith since λ/l u 1. ρ ɛ = 2µ s ij s ij ρ U 3 l u, or ɛ U 3 /l u, so that This implies a large scale separation The dissipation rate ɛ is approximately independent of viscosity (counter-intuitive. This is one aspect of Reynolds number similarity. And, ɛ U 2 /t u, where t u = l u /U, i.e., the time scale of energy decay or energy transfer is l u /U. The modeling must be consistent with this. It is often convenient to define the following length scale η, ( ν 3 1/4 η =, the Kolmogorov length scale. (26 ɛ Later in the course it will be shown that η is the scale at which viscous effects become of order 1, and that for length scales much larger than η, the viscous effects are negligible. From the results of scaling the turbulent kinetic energy equation, we can estimate the other ratios of the various length scales. For example, ( η ν 3 = l u ɛl 4 u η l u = O 1/4 [( ν 3 l u = O U 3 l 4 u ( Re 3/4 l 1/4 ], or as Re l. (27 This is an important result, and tells us that the bandwidth of the turbulence increases as Re 3/4 l as Re l increases. Higher Reynolds number flows have higher bandwidth. With this equation and the result for λ/l u above, then ( η λ = O Re 1/4 l as Re l. (28 Another useful Reynolds number to consider is the Taylor Reynolds number, R λ, which is defined by R λ = Uλ ν. (29 This is often measured in turbulence experiments since λ can be measured much more accurately that an integral scale, say L 11, which is a useful measurement of the large, energy-containing scales of turbulence. Note that R l = Uλ ν = Ul u ν ( λ lu ( = Re l O Re 1/2 l (, so that R λ = O So Equation 28 can be rewritten for η/λ in terms of R λ as η ( λ = O R 1/2 λ as Re l. Re 1/2 l as Re l. These results lead to the following intuitive picture of turbulence as a function of length scale, as depicted in Figure 5. The energy is input at the scale l u, which depends on the input method, 12

13 Figure 5: Sketch of relationship between important scales in a turbulent flow and energy transfer. e.g., flow instabilities, a pipe diameter, a jet outlet diameter, etc. This energy is then transferred to intermediate length scales, of the order of the Taylor scale, due to processes like vortex stretching, the transfer rate depending on U and l u, and the time scale of transfer being l u /U. Energy continues to be transferred downscale to the Kolmogorov scale η, where it is dissipated into heat at the rate of ɛ. The energy transfer rate to the Kolmogorov scale is the same as the dissipation rate itself. From these ideas comes the argument that the Kolmogorov scale motions adjust to dissipate all of the energy that is transferred to them, and that what happens at these scales does not directly affect the large scale motions. Figure 6: Sketch of a typical turbulence energy spectrum including the energy-containing (input range, typified by 2π/l u, and the dissipation range, typified by 2π/η. Note at as the Reynolds number is increased with l u fixed, 2π/η will move to the right. Another way to view this is in terms of energy spectra; a typical turbulence energy spectrum is depicted in Figure 6. Most of the energy in the flow is at the large scales, which are of the order of 2π/l u in the plot. The energy is dissipated at the scales of the order of 2π/η. If the Reynolds number of the flow were increased, the bandwidth of the plot would increased which, for fixed l u, would result in the Kolmogorov wave number moving to the right. This would not affect, however, the motions at the large scales. This idea is the essence of Reynolds number similarity, which is discussed in the next section. 13

14 1.7 Reynolds Number Similarity Applied to Turbulent Jets Figure 7 below shows three different turbulent jets at vastly different Reynolds numbers (the figure is from Paul Dimotakis at Caltech. The three cases shown are: a liquid-phase turbulent jet with R D (flow A; a liquid-phase turbulent jet with R D 10 4 (flow B; and a rocket test with R D > 10 8 (flow C. (The Reynolds number R D will be defined below. The jets appear to be qualitatively very similar, although it is clear that the overall flows are very different in terms of flow speed, jet diameter, and fluid. It is claimed that, due to Reynolds number similarity, the statistical properties of the three flows can be the same, even though the speeds, diameters, and fluids, and hence the Reynolds numbers, were different. The following contains more explanation of these comments. Figure 7: Liquid-phase turbulent jet, R D (flow A; liquid-phase turbulent jet, R D 10 4 (flow B; rocket test, R D > 10 8 (flow C Start by assuming each flow obeys the Navier-Stokes equations and is in incompressible (isochoric motion, i.e., in vector notation ρ ˆt Û + ρ(û ˆ Û = ˆ ˆP + µ ˆ 2 Û ˆ Û = 0. The equations are dimensional, and ( denotes a dimensional quantity. In order to nondimensionalize the equations, for a velocity choose the maximum outflow velocity of the jets, say U D (which is different for the three flows. Therefore a nondimensional velocity, U, is given by U = Û/U D. For the length scale choose the jet diameter D (which is also different for the three cases. Finally, choose the time scale to be L/U D. When these variables are used to nondimensionalize the equations, with the pressure nondimensionalized as P = ˆP /ρ UD 2, the following equations are obtained: 1 U + (U U = P + 2 U t R D 14

15 U = 0. Here R D = U D D/ν, a Reynolds number characterizing each of the jet flows. Note that, as mentioned above, R D is different for the three flows, hence the governing equations are different, and so one might expect that the statistical characteristics of these flows would be different as well. Consider, however, the nondimensional equations for the mean flow, i.e., in index notation (with the viscous terms expressed in terms of the average strain rate: t U i + U j x j U i = x i P + x i U i = 0. x j { 2 1 } S ij u i u j R D Note that, as argued prreviously, for high Reynolds number flows it is expected that the viscous terms in the equation for the mean flow are unimportant. If they are neglected, then the Reynolds number dependence of the equations (involving R D is lost. Furthermore, the nondimensional boundary conditions for the three flows are the same, expressed in terms of the nondimensional inlet velocity. (It is assumed as well that the nondimensional jet outlet profiles are the same, although it can be argued that this requirement is not important. Therefore we would expect that the properties of nondimensional velocity U i should be the same for all three flows, expressed in terms of the nondimensional distance x = ˆx/D. This implies that is the same for all three flows, so that [ ] Û i U D U i (x = Ûi U D (Dx A [ ] Û i = U D B [ ] Û i = where the subscripts A, B, and C refer to the flows defined above, and the dependence on ˆx is not shown. Similar arguments hold for the kinetic energies as well, where again for high Reynolds numbers the terms involving viscous transport (in terms of R D can be neglected. In the turbulent kinetic energy equation, however, the dissipation-rate term is assumed to be independent of the Reynolds number and must be retained, as discussed in the previous subsection. An implication of the Reynolds-number independence of the dissipation rate terms is that [ ] [ ] [ ] 2µ ŝij ŝ ij D 2µ ŝij ŝ ij D 2µ ŝij ŝ ij D ɛ A = = ɛ B = = ɛ C =, U 3 D A where again the ˆx dependence is suppressed. Therefore, for example, ŝ ij ŝ ij A = ( UDA U DB U 3 D 3 ( DB µ B D A µ A U D B C, ŝ ij ŝ ij B. Note that conclusions regarding Reynolds-number independence apply to statistical properties which depend on the large-scale features of the flow field, such as U i, u i u j, and ɛ. Statistical properties of the flow which depend on its small scale features, such as the Taylor microscale λ and the mean square vorticity ˆω i 2, are expected to depend on the Reynolds number. For example, as will be proven later, the dissipation rate can be written approximately in terms of the mean square vorticity as: ˆɛ = 2ν ŝ ij ŝ ij = ν ˆω i 2 [1 + O(RD 1 ], 15 U 3 D C

16 taking U D to be the characteristic turbulence velocity scale, and D the characteristic length scale. Then with the nondimensionalization used above for the dissipation rate, ɛ = D ˆɛ U 3 D ( Dν ˆω 2 = i U 3 D [1 + O(R 1 D ] = ν U D D ω2 i [1 + O(R 1 D ], with the mean-square vorticity nondimensionalized as ωi 2 = D2 ˆω i 2 UD 2. Therefore, ɛ = ω2 i R D [1 + O(R 1 D ]. Solving for ω 2 i, ω 2 i = R D ɛ + O(1. Therefore the mean square fluctuating vorticity depends on the Reynolds number and is, for large R D, proportional to R D. 2 Simple shear flow Figure 8: Simple shear flow with mean velocity in the x 1 direction, and non-homogeneity and mean momentum and energy fluxes in the x 2 direction. It could be, for example, a temporally evolving turbulent shear layer or planar turbulent jet, both often studied using direct numerical simulation. Consider the simplified turbulent flow which has the following propertries (see Figure 8: U = [ U 1 (x 2, 0, 0]. High Reynolds number (i.e., R l, so the term 2µ S ij can be neglected in the x j { } averaged momentum equation, 2µ U i S ij in the mean flow energy equation, and x 2 {2µ u j s ij } in the turbulence kinetic energy equation. x j Homogeneous in x 1 and x 3, so that x 1 ( = 16 x 3 ( = 0.

17 Such a flow has been set up in the laboratory, and on the computer (see the text, section The mean momentum equation reduces to: ρ t U 1 = x 2 ρ u 1 u 2. So the momentum in the x 1 direction only changes due to turbulent transport in the x 2 direction. Integrating this equation in x 2 gives: d dt ρ U 1 (x 2 dx 2 total x 1 momentum = ( ρ u 1 u 2 x 2 dx 2 = ρ u 1 u 2 = 0, if u 1 u 2 0 as x 2 ±, i.e., assuming that the turbulence is localized in x 2. So the overall momentum is conserved, since there are no external forces acting on the flow. But the mean velocity profile is broadening due to turbulence transport. The mean flow energy equation reduces to, for this problem: t ρ U ( = ρ U 1 u 1 u 2 + ρ u 1 u 2 U 1. x } 2 x {{}}{{ 2 } transport by turbulence loss to turbulence Therefore, although the overall mean momentum is conserved, the overall mean flow kinetic energy is not. Note that, for this particular case, using a turbulence viscosity model for the Reynolds stress, i.e., u 1 u 2 = 2ν T S 12 = ν T U 1, then x 2 ρ u 1 u 2 ( U1 2 U 1 = ρν T < 0, assuming ν T > 0. x 2 x 2 Therefore, for the turbulence viscosity model, energy always goes from the mean flow to the turbulence. Furthermore, the viscous dissipation rate reduces to ( U1 2 2µ S ij S ij = ρν, and x 2 ( U1 ρν viscous dissipation rate loss rate to turbulence = x 2 ρν T ( U1 x = ν ν T 1 Re l with ν T Ul u. It is of interest to examine each component of the turbulence energy equation. The equations for each can be obtained as follows: Obtain an equation for u i (i = 1, 2, 3 by subtracting the equation for U i from the equation for U i, and multiplying the resulting equation for u i by u i (no sum on i and averaging. This gives: 17

18 p using u 1 = x 1 1 t 2 ρ u2 1 = ρ u 1 u 2 U 1 + x 2 1 t 2 ρ u2 2 = t 2 ρ u2 3 = 0 + x 1 u 1 p p u 1 x 1 p u 2 x 2 p u 3 x 3 u 2 ρ u2 1 2 ( u 2 (p + ρ u2 2 x 2 x 2 x 2 u 2 ρ u ρɛ 2 ρ 1 3 ɛ ρ 1 3 ɛ (30 + p u 1, and similarly for the u 3 component. It has also x 1 =0, homog. in x 1 been assumed for simplicity that the dissipation rate is approximately isotropic, which will be explained later in the course. Note that p u 1 is a pressure/strain-rate correlation; its modeling x 1 is a key component to modeling the Reynolds stress equations. Also note that the turbulence kinetic energy equation reduces to, for this problem: 1 t 2 ρ u2 i = ρ u 1 u 2 U ( u 2 (p + ρ u2 i x 2 x 2 2 ρɛ. These results imply the following. 1. The turbulence production occurs entirely in the equation for u 2 1, i.e., in the component corresponding to the mean flow. The other equations have no production terms. 2. The transport terms only redistribute energy in space, in this case only in the non homogeneous direction x The dissipation terms act only to drain energy out of the turbulence. 4. The pressure/strain rate terms sum to 0, i.e., p u 1 + p u 2 + p u 3 = p u i = 0, x 1 x 2 x 3 x i and hence to not appear in the turbulence kinetic energy equation. 5. The pressure/strain rate terms only act to exchange energy among the three components, without changing the total energy (an important fact in modeling. 6. u 2 2 and u2 3 must receive energy entirely from the pressure/strain-rate terms. 7. If u 2 2 and u2 3 are to be maintained, which they are observed to be experimentally, then p u 1 < 0; p u 2, p u 1 > 0. A model due to Rotta, and its variations, assures this x 1 x 2 x 1 behavior. For a physical interpretation, refer to Figure 9. The flow is converging in the x 1 direction, i..e., u 1 < 0, which leads to a local pressure increase, that is p > 0. So p u 1 < 0 x 1 x 1 locally, leading to a loss in u 2 1, and a gain in u2 2 and u In most shear flows, with the mean flow mainly in the x 1 direction, say, then it is found that u 2 1 is roughly a factor of 2 larger than u2 2 and u Neglecting terms of order 1/Re l in the equations for u 2 1, u2 2, and u2 3, then the mathematical problem no longer depends upon Re l (or viscosity, i.e., there is Reynolds number similarity. 18

19 Figure 9: Flow with u 1 x 1 < 0 and p > 0 locally. 3 Free Shear Flows (Boundary-Free Flows Note that there are a number of photographs, videos and animations on free shear flows in Van Dyke s book (especially pages , in the Turbulence module in Multimedia Fluid Mechanics II, and in Robert Stewart s film Turbulence. 3.1 Background It is usual in discussing turbulent flows to distinguish between: flows that are directly influenced by a boundary, such as boundary layer flows, flows around an object, pipe and channel flows, etc., and free-shear flows, where boundaries are of influence, at most, in the initiation of the flows, e.g., jets, wakes, and mixing layers. Note that flows are also distinguished in this manner when considering the stability of laminar flows. There are several reasons that this distinction is made. 1. For high Reynolds number flows, free-shear flows can become effectively independent of the Reynolds number. For boundary layer flows, as the boundary is approached, U and l u become sufficiently small that Re l = Ul u /ν 1 locally, so that the flow does depend on the Reynolds number. 2. The dynamics are much different due to the boundary. For example, vorticity is produced at the boundary, and there can therefore be very high vorticity very near it; flow instabilities there produce particular type of streaks and bursts which are not observed in free-shear flows; free-shear flows are susceptible to larger-scale instabilities which appear to produce coherent-type structures; because of their stability characteristics, free-shear flows are very responsive to forcing. 19

20 3. Pressure fluctuations are strongly influenced by a boundary. 4. The boundary has a strong non-isotropizing effect. Here we will consider the example of turbulent wakes, which have most of the properties of other free shear flows. 3.2 Turbulent wakes some characteristics Consider the sketch of a turbulent wake of a circular cylinder given in Figure 10. Assume that the flow is moving steadily past the cylinder of diameter D at a steady speed of U 0. The wake has a number of characteristics which are important to consider, some which are the following. Figure 10: Sketch of the turbulent wake of a circular cylinder. 1. The wake is laminar and steady for, approximately, R D = ρu 0 D/µ Vortex shedding occurs in the range 60 R D 5000, with the Strouhal number S given by S = nd/u , where n is the frequency of vortex shedding. 3. For R D > 5000, the wake is strongly turbulent, but vortex shedding is still important. 4. The flow in the wake region is highly turbulent and rotational; outside the wake region the flow is highly unsteady but irrotational. A thin layer called the superlayer separates the turbulent and non-turbulent regions. 5. As it evolves downstream, the wake spreads by entraining previously non-turbulent fluid. 3.3 Turbulent wake analysis Consider the mean velocity profile in the wake of a circular cylinder, shown in Figure 11. Assume a Cartesian coordinate system with x in the flow direction, y normal to the flow direction and the axis of the cylinder, and z along the cylinder axis; the corresponding velocity components are (U, V, W. The ambient flow speed is U 0, and the mean velocity in the x-direction is U. Of particular interest, shown in the figure, is the velocity deficit U d = U 0 U, and the average velocity deficit U d = U 0 U. The following assumptions are made: 20

21 Figure 11: Velocity profiles of an evolving wake. Note the distinction between the mean axial velocity U (x, y and the mean velocity deficit U d (x, y = U 0 U (x, y. 1. Statistically steady, i.e., ( = 0. t 2. Statistically homogeneous in the cross-stream (z direction, i.e., ( = 0. That is, we are z dealing with the plane wake of, effectively, a two-dimensional object. 3. No cross-stream mean velocity, i.e., W = High Reynolds number flow (to be defined below. 5. Almost parallel flow, i.e., ( (. That is, if l is a characteristic differential x y scale of the mean flow in the y-direction, and L is a characteristic differential scale in the x-direction, then l/l 1. This is a boundary layer-type assumption, and is usually true for high Reynolds number flows at a downstream distance of about 20D or more, i.e., x/d > 20. We will therefore be examining the far field behavior of the wake, which will teach give much information about free-shear flows Scaling the Mean Flow Equations Consider the two-dimensional, incompressible, turbulent wake of a cylinder, and described by the time-averaged Navier-Stokes equations: uv (31 U U + V U x y U V + V V x y U x + V y = 1 ( p 2 ρ x + ν V = 1 ρ x V y 2 ( p 2 y + ν V x V y 2 x u2 y x uv y v2 (32 = 0. (33 As mentioned above, the ambient flow is in the x-direction, with y transverse to the flow. Furthermore, the flow has been assumed to be statistically steady and statistically homogeneous in the z-direction. 21

22 Taking U 0 to be the (constant ambient flow speed, the following two lengths scales are defined: L the length in the x-direction over which the streamwise mean velocity deficit, U 0 U, changes by a specified amount, say decreasing by a factor of 2; l the length in the y-direction over which U 0 U changes by the same amount. It is observed experimentally that l/l 1, i.e., the velocity changes much more quickly in y than in x, i.e., it is thin. The following rough scaling analysis is based upon two principal assumptions: 1. l(x/l(x 1, i.e., the wake is locally (in x thin, and 2. Re = U s (xl(x/ν 1, i.e., the local (in x Reynolds number of the flow is very high, where, over all values of y at a given downstream location x, U s (x = max[u 0 U ] is the peak mean streamwise velocity deficit. A rough estimate will be made for each term in the time-averaged equations. To do this, an order-of-magnitude estimate for each dependent and independent variable will be needed. These estimates are as follows. U d U s. Here the symbol implies order-of-magnitude equality, and denotes a differential scale. U 0 is the advective scaling for U. V, V V s =?. That is, the scaling for the y-component of the mean velocity, V, is not known at this point. x L y l u 2, uv, v 2 u s 2 p ρu 2 s, a dynamic pressure, using a Bernoulli-like scaling for pressure. From laboratory data, it is found that u s U s, so the two will be taken as equal here, and only U s will be used below. Using this scaling in the continuity equation, Equation (33, gives V s l U s, or (34 ( L l V s U s U s (35 L i.e., V s is expected to be very small compared to U s, which is consistent with the assumption that the wake is thin. Next, scaling is presented for each term in the y-component of the mean momentum equation, Equation (32: U V x U 0 U sl/l 2 (U 0 /U s(l/l 2 + V V y U 2 s l 2 /L 2 l (l/l 2 = 1 p ρ y U 2 s /l 1 ( 2 V +ν } x {{ 2 } νu sl/l 3 (ν/u sl(l/l V y }{{ 2 } νu s/ll (ν/u sl(l/l x uv U 2 s /L (l/l y v2 U s 2 /l 1 (36

23 In the above equation, the first line below the equation represents the scaling of each term, while the second line represent the scaling, but divided by the common term U 2 s /l. From the assumptions above, only the first term on the LHS, and the first and last terms on the RHS will possibly remain after neglecting the other terms, which are very small by assumption. It will be established below that the first term on the LHS can also be neglected. Therefore, the y-component of the mean momentum equations reduces to: 0 = 1 p ρ y y v2, or 0 = y Integrating Equation (37 in y for a fixed x then gives: ( p ρ + v2. (37 p ρ (x, y + v2 (x, y = P o(x, (38 ρ where P o (x is the ambient pressure (in the far field, since v 2 0 as y ±, i.e., the turbulence decays to 0 far from the wake. In the discussion that follows we will assume the ambient pressure is constant in space, i.e., P o (x = P o = constant. Therefore, differentiating Equation (38 in x gives, in addition to Equation (37 above: Therefore p /ρ + v 2 is uniform in space. ( p x ρ + v2 = 0. Next the x-component of the mean momentum equation is scaled in the same way, giving: U U x (U 0 U s/l (U 0 /U s(l/l + V U y (U 2 s /L (l/l = 1 p ρ x U 2 s /L l/l ( 2 U +ν } x {{ 2 } νu s/l }{{ 2 } (ν/u sl(l/l U y }{{ 2 } νu s/l }{{ 2 } (ν/u sl x u2 U 2 s /L (l/l y uv U s 2 /l 1 Again the first line below the equation represents the scaling of each term, while the next line represents the scaling, but divided in this case by the common term Us 2 /l. From the original assumptions, the only possible terms of order 1 are the first term on the LHS and the last term on the RHS. It is expected that the momentum equation predicts the evolution of U in the downstream direction, so the first term on the LHS will be important. The last term on the RHS must be important or there will be no turbulence effects on this flow. The fact that these two terms are in balance then implies, in particular, that U 0 l U s L 1, or U s l U 0 L. Using this estimate of U 0 /U s in the first term of the LHS of Equation (32 implies that it is of order l/l compared to the last term on the RHS, and can therefore be neglected, as assumed. The x-component of the mean momentum equation therefore reduces to: (39 U U x = uv (40 y 23

24 Since U 0 is constant, this equation can be written as: U x ( U U 0 = U x ( U d = uv, (41 y where, as discussed above, the mean velocity deficit U d = U 0 U scales as U s. Therefore the advective velocity U in Equation (40 can be written as: U = U 0 (U 0 U = U 0 U d = U 0 (1 + O(U s /U 0 = U 0 (1 + O(l/L, so that U x can be approximated as U 0 in Equation (40. x Therefore, based upon the assumptions that l/l 1 and Re 1, the x-component of the mean momentum equation and the continuity equation reduce to: U U 0 x = uv (42 y U x + V y = 0. (43 These equations are often referred to as the boundary layer equations, although they apply to jets and wakes as well as boundary layers. For the wake they are a fairly good approximation away from the near field, for about x/d > 10 20, depending on the shape of the object. Note that there is still a closure problem, since there are two equations but with three unknowns, U, V, and uv. Therefore a closure assumption is still needed in order to utilize these equations Momentum flux deficit Next, multiplying Equation (42 by ρ and writing it in terms of the momentum deficit gives, since U 0 is a constant: U 0 ρ x [U 0 U ] = ρ uv. (44 y For fixed x, integrating this from y = to + gives: U 0 ρ d dx x [U 0 U ]dy = ρ } { ρ U 0 [U 0 U ]dy uv dy, or (45 y = ρ uv = 0, (46 using Leibnitz rule for commuting the x-derivative and the integral, realizing that the integrand on the RHS contains an exact differential, and assuming that the Reynolds stress ρ uv decays sufficient fast as y ±. Therefore the momentum deficit flux, M, defined by M ρ U 0 [U 0 U ]dy, (47 is a constant, independent of x. This implies that the mean velocity U cannot evolve downstream in x in an arbitrary fashion, but must conserve the momentum deficit flux. The reason for this is that no overall forces are applied to the late wake. But, for example, if an ambient pressure gradient existed, i.e., dp o dx 0, then in general M would change with x. 24

25 The momentum deficit flux is the principal feature that defines a wake, and can be related to the drag force on the cylinder. Also, a length scale called the momentum thickness, Θ, is defined in terms of M as, ρu0 2 Θ = M, or Θ = ( 1 U U Similarity analysis for the turbulent wake dy. (48 When measurements of the mean velocity are made in the wake, at least for about x/d > 20 or so, it is found that the mean axial velocity profile evolves in the downstream direction, with the peak velocity decreasing as the wake broadens (see Figure 12. The profile does not evolve in an arbitrary manner, however. When the data are plotted with U d normalized by the local (in x peak velocity, U s, and the y distance normalized by a local (in x length scale characterizing the wake width, l, to be defined below, the curves collapse as shown in the figure. This implies that, approximately, Figure 12: Measurements of the mean axial velocity at x 1, represented by the symbol x, and at x 2, represented by the symbol o. To the right both cases are plotted in similarity form, with U d (x, y plotted versus y/l(x. U s (x U d (x 1, y U s (x 1 [ y ] = f, and U d (x 2, y [ y ] = f, l(x 1 U s (x 2 l(x 2 where f is a function to be determined. We assume more generally that the mean wake deficit U d (x, y can be written in this form for any x, i.e.: U d (x, y U s (x [ ] y = f, (49 l(x where l(x is a measure of the wake width, to be determined. Flows having this characteristic are called self-similar or self-preserving. They have attained a steady-state when case in proper terms. Implicit is that there is only one velocity scale and one length scale characterizing the flow. A number of other flows, e.g., boundary layers, jets, and mixing layers can be self-similar. 25

26 The momentum and continuity equations, Equations (42 and (43 imply that V and uv should also be written in similarity form, i.e., [ ] V (x, y y = h, and (50 U s (x l(x uv [ ] y Us 2 (x = g. (51 l(x where h and g are also to be determined. These similarity forms will be next used in Equation (44, a reduced form of the x-component of the mean momentum equation. Note that, using the similarity form for U d, and chain rule in calculus, x U d (x, y = ( U s (xf(y/l(x = U x sf f η(l /lu s, (52 where η = y/l(x, and ( denotes a derivative of a function with respect to its argument. Furthermore, y uv = ( Us 2 (xg(y/l(x = U 2 1 s y l g. (53 Using Equations (52 and (53 in Equation (44 then gives: { } U 0 U sf U s f η l = U 2 1 s l l g, or, multiplying through by l/us 2, ( U ( U s l 0 Us 2 l f + U 0 U s f η = g. (54 Note that the RHS, g is a function only of η. Each term on the LHS, however, separates into a function of η times a function of x. Since the coefficient of g is independent of x, then the coefficients of the terms f and f η on the LHS must also be independent of x in order for the equation to be consistent. Therefore, U s Us 2 l = c 1, a constant, and (55 l U s = c 2, another constant. (56 It will be assumed that both U s and l have algebraic dependence on x, i.e., l(x x n, and U s (x x m. Note that if these assumptions are erroneous, then inconsistencies will arise is the following analysis. Plugging the first of these into Equation (55 and only considering the x-dependent part of the result gives: x m 1 x 2m xn = constant = x 0, so that m 1 2m + n = 0, or m = n 1. Plugging the same expressions into Equation (56 will give the same result. Therefore, l(x and U s (x must satisfy l(x x n, and U s (x x n 1. 26